# Variational formulation of harmonicity on Riemannian manifolds

$\newcommand{\R}{\mathbb{R}}$ I am trying to follow a derivation of the first variation formula for the energy functional. (In "Selected Topics in Harmonic maps").

Here is the context:

$M,N$ are Riemannian manifolds, $\phi:M \to N$ is a smooth map which is a critical point of the energy functional $\int_M |d\phi|^2$ where $|\cdot|$ is the natural induced norm (inner product) on $T^*M \otimes \phi^*(TN)$.($\phi^*(TN)$ is the pullback bundle of $TN$ along $\phi$)

The author then defines a family of maps $\phi_t:M \to N$ ($\phi_0=\phi$),

i.e a map $\psi:M \times \R \to N \, , \, \psi(p,t)=\phi_t(p)$

$$0=\frac{1}{2}\frac{d}{dt}\int_M |d\phi_t|^2 \text{vol}_g = \frac{1}{2}\int_M \frac{\partial}{\partial t}\langle d\phi_t,d\phi_t \rangle \text{vol}_g \stackrel{(*)}{=} \int_M \langle \nabla_{\frac{\partial}{\partial t}} d\phi_t,d\phi_t \rangle \text{vol}_g$$

(where all the derivatives are taken at time $t=0$)

Here is my problem in a nutshell:

The metric and connection in the two sides of $(*)$ are not defined on the same bundles.

Details:

The author mentions that the connection $\nabla$ in the formula above, is the induced one on $T^*(M \times \R)\otimes \psi^*(TN)$

I guess equality $(*)$ follows from metricity (All the original connections are Levi-Civita, hence they and everything induced by them is metric).

But this metricity is w.r.t the metric on $T^*(M \times \R)\otimes \psi^*(TN)$, hence instead of $\langle d\phi_t,d\phi_t \rangle$ I think it's supposed to be written $\langle d\psi,d\psi \rangle$, where this is the (squared) norm of the differential of the map $\psi$, i.e at each point $(p,t)$ this is the induced norm on $\operatorname{Hom}(T_{(p,t)}(M \times \R),T_{\phi_t(p)}N)$

(This is the only way I am able to make sense of this. I am assuming that $\R$ is endowed with its standard metric, and $M \times \R$ with the product metric).

The problem is that now there is no reason this should be zero, since the variational assumption is that $\frac{d}{dt}\int_M |d\phi_t|^2 \text{vol}_g=0$ , not $\frac{d}{dt}\int_M |d\psi|^2 \text{vol}_g=0$, and I see no apparent reason why it should hold that $|d\psi|=|d\phi_t|$ in general.

Am I missing something? what is going on here?

If $E \to X$ is a vector bundle over a manifold $X$ with metric $\langle -,-\rangle$ and connection $\nabla$, metric compatibility is the condition $$\xi \langle s_1, s_2 \rangle = \langle \nabla_\xi s_1, s_2 \rangle + \langle s_1, \nabla_\xi s_2 \rangle$$ for any vector field $\xi$ on $X$ and any two sections $s_1, s_2$ of $E$.
In your case, we can think of $d\phi$ (with $t$ varying) as a section of the bundle $p^\ast(T^\ast M) \otimes \psi^\ast(TN)$ over $M \times \mathbb{R}$, where $p: M \times \mathbb{R} \to M$ is the projection onto the first factor. This bundle has a natural metric and connection induced by those on $TM$ and $TN$. This connection is compatible with the metric, so for any vector field $\xi$ we have $$\xi |d\phi|^2 = 2 \langle \nabla_\xi d\phi, d\phi \rangle$$ Take $\xi = \partial / \partial t$ to obtain (*).
Theorem Let $$\phi_t$$ be a variation of $$\phi=\phi_0: M \to N$$ with variational field $$\upsilon$$. Then \begin{align*} \dfrac{d}{dt} E(\phi_t) \bigg|_{t=0}= -\int_M\langle \upsilon, \tau(\phi)\rangle d\mu_M + \int_{\partial M} \left\langle \upsilon, \ d\phi(N)\right\rangle d\mu_{\partial M}. \end{align*} where $$\tau(\phi) = \text{tr} \nabla^{T^*M\otimes \phi^*TN} d\phi = -d^*_\nabla d\phi$$, $$\mu_M$$ and $$\mu_{\partial M}$$ are Riemannian densities on $$M$$ and $$\partial M$$ respectively, and $$N$$ is the out-pointing normal on $$\partial M$$.
Remark. The operator $$d_{\nabla}^*: \Omega^\bullet(M; \phi^*TN) \to \Omega^{\bullet-1}(M; \phi^*TN)$$ is the formal adjoint of the exterior differential operator $$d_\nabla: \Omega^\bullet(M; \phi^*TN) \to \Omega^{\bullet+1}(M; \phi^*TN)$$ with respect to $$L^2$$ inner product, that is \begin{align*} \int_M \left\langle d_\nabla \omega, \eta\right\rangle_{\Omega^p(M;\phi^*TN)} = \int_M\left\langle \omega, d^*_\nabla \eta\right\rangle_{\Omega^{p-1} (M;\phi^*TN)}, \quad \forall \omega \in \Omega^{p-1}, \, \eta \in \Omega^p. \end{align*} For details see page 7-9 of Eells & Lemaire's Selected Topics in Harmonic Maps.
Proof. Let $$\Phi: M\times I \to N$$ be the variation defined by $$\phi_t(x) = \Phi(x, t)$$, then \begin{align*} \dfrac{d}{dt}E(\phi_t) = \dfrac{1}{2}\dfrac{d}{dt}\int_M \left\langle d\phi_t \ , \ d\phi_t\right\rangle =\dfrac{1}{2} \int_M \dfrac{\partial } {\partial t}\left\langle d\phi_t \ , \ d\phi_t\right\rangle \end{align*} We verify the equation \begin{align*} \dfrac{1}{2}\dfrac{\partial }{\partial t}\left\langle d\phi_t \ , \ d\phi_t\right\rangle_{T^*M\otimes\phi_t^*TN} = -\left\langle \partial_t\Phi \big|_t\ , \ \tau(\phi_t)\right\rangle_{\phi_t^*TN} + \ \text{div} \, \theta_t^\sharp \end{align*} where $$\partial_t\Phi \big|_t = d\Phi(\partial_t)(\boldsymbol{\cdot},t)$$ is a section of $$\phi_t^*TN$$, and $$\theta_t \in \Omega^1(M)$$ is defined by \begin{align*} \theta_t(X) = \left\langle \partial_t \Phi|_t, \ d\phi_t(X)\right\rangle_{f_tT^*TN}. \end{align*} For doing this, we compare them at a fixed point $$p$$, centered at which normal coordinates $$x^i$$ are chosen. We use the notation $$\partial_i \Phi$$ for the section $$d\Phi(\partial_i)$$ of $$\Phi^*TN$$. By the metricity of the connection on $$\Phi^*TN$$, \begin{align*} \dfrac{\partial }{\partial t}\left\langle d\phi_t \ , \ d\phi_t\right\rangle_{T^*M\otimes\phi_t^*TN}(p) &= \dfrac{\partial } {\partial t}\left\langle d\phi_t(\partial_i) \ , \ d\phi_t(\partial_i)\right\rangle_{\phi_t^*TN}(p,t)\\ &=\dfrac{\partial }{\partial t}\left\langle \partial_i\Phi \ , \ \partial_i\Phi \right\rangle_{\Phi^*TN}(p,t)\\ &=2\left\langle \nabla^{\Phi^*TN}_{\partial_t}\partial_i\Phi \ , \ \partial_i\Phi \right\rangle_{\Phi^*TN}(p,t). \end{align*} Since \begin{align*} \nabla^{\Phi^*TN}_{\partial_t} \partial_i\Phi =\nabla^{\Phi^*TN}_{\partial_i} \partial_t\Phi - d\Phi[\partial_i, \partial_t] =\nabla^{\Phi^*TN}_{\partial_i} \partial_t\Phi, \end{align*} we conclude that \begin{align*} \dfrac{1}{2}\dfrac{\partial}{\partial t}\left\langle d\phi_t \ , \ d\phi_t \right\rangle_{T^*M\otimes\phi_t^*TN}(p, t) &= \left\langle \nabla^{\Phi^*TN}_{\partial_i} \partial_t\Phi \ , \ \partial_i\Phi \right\rangle_{\Phi^*TN}(p,t)\\ &=\left\langle \nabla^{\phi_t^*TN}_{\partial_i} \partial_t\Phi \big|_t\ , \ d\phi_t(\partial_i) \right\rangle_{\phi_t^*TN}(p)\\ &=\left\langle \nabla^{\phi_t^*TN} \partial_t\Phi \big|_t\ , \ d\phi_t\right\rangle_{T^*M \otimes \phi_t^*TN}(p)\\ &=\left\langle d_\nabla \partial_t\Phi \big|_t\ , \ d\phi_t\right\rangle_{T^*M \otimes \phi_t^*TN}(p)\\ &=\left\langle \partial_t\Phi \big|_t\ , \ d_\nabla^*d\phi_t\right\rangle_{\phi_t^*TN}(p)+\left(\text{div} \, \theta_t^\sharp\right)(p)\\ &=-\left\langle \partial_t\Phi \big|_t\ , \ \tau(\phi_t)\right\rangle_{\phi_t^*TN}(p)+\left(\text{div} \, \theta_t^\sharp\right)(p). \end{align*} For the second last equality check Proposition 2.2.7 and Proposition 2.2.8 of Petersen's Riemannian Geometry, 3rd Edition.